Type The Correct Answer In Each Box. Use Numerals Instead Of Words.Consider The Equation:${ \frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3} }$The Equation Has $\square$ Valid Solution(s) And $\square$ Extraneous Solution(s).

The Equation and Its Solutions

Introduction

When dealing with equations involving fractions, it's essential to find a common denominator to simplify the equation and solve for the variable. In this case, we have an equation with three fractions, and our goal is to find the number of valid and extraneous solutions.

The Equation

The given equation is:

$$\frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}$$

To solve this equation, we need to find a common denominator for the fractions. The least common multiple (LCM) of $x-3$ and $x$ is $x(x-3)$.

Multiplying Both Sides by the Common Denominator

We multiply both sides of the equation by $x(x-3)$ to eliminate the fractions:

$$2x + (x-3) = (x-1)x$$

Expanding and Simplifying

Now, we expand and simplify the equation:

$$2x + x - 3 = x^2 - x - 1$$

Combine like terms:

$$3x - 3 = x^2 - x - 1$$

Rearranging the Equation

We rearrange the equation to get a quadratic equation in standard form:

$$x^2 - 4x + 2 = 0$$

Solving the Quadratic Equation

To solve the quadratic equation, we can use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = -4$, and $c = 2$. Plugging these values into the formula, we get:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$$

Simplifying the expression under the square root:

$$x = \frac{4 \pm \sqrt{16 - 8}}{2}$$

$$x = \frac{4 \pm \sqrt{8}}{2}$$

$$x = \frac{4 \pm 2\sqrt{2}}{2}$$

$$x = 2 \pm \sqrt{2}$$

Valid and Extraneous Solutions

We have two possible solutions: $x = 2 + \sqrt{2}$ and $x = 2 - \sqrt{2}$. To determine which of these solutions are valid and which are extraneous, we need to check if they satisfy the original equation.

Checking the Solutions

We substitute $x = 2 + \sqrt{2}$ into the original equation:

$$\frac{2}{(2 + \sqrt{2}) - 3} + \frac{1}{2 + \sqrt{2}} = \frac{(2 + \sqrt{2}) - 1}{(2 + \sqrt{2}) - 3}$$

Simplifying the expression:

$$\frac{2}{-1 + \sqrt{2}} + \frac{1}{2 + \sqrt{2}} = \frac{1 + \sqrt{2}}{-1 + \sqrt{2}}$$

This expression is undefined, so $x = 2 + \sqrt{2}$ is an extraneous solution.

Checking the Second Solution

We substitute $x = 2 - \sqrt{2}$ into the original equation:

$$\frac{2}{(2 - \sqrt{2}) - 3} + \frac{1}{2 - \sqrt{2}} = \frac{(2 - \sqrt{2}) - 1}{(2 - \sqrt{2}) - 3}$$

Simplifying the expression:

$$\frac{2}{-1 - \sqrt{2}} + \frac{1}{2 - \sqrt{2}} = \frac{1 - \sqrt{2}}{-1 - \sqrt{2}}$$

This expression is also undefined, so $x = 2 - \sqrt{2}$ is an extraneous solution.

Conclusion

We have found that both solutions, $x = 2 + \sqrt{2}$ and $x = 2 - \sqrt{2}$, are extraneous solutions. Therefore, the equation has 0 valid solution(s) and 2 extraneous solution(s).

Discussion

This problem requires careful manipulation of the equation and attention to detail when checking the solutions. The use of the quadratic formula and the simplification of the expressions under the square root are essential steps in solving the equation. The final answer is a reflection of the importance of checking the solutions to ensure that they satisfy the original equation.

Q&A: Understanding the Equation and Its Solutions

Introduction

In the previous section, we solved the equation $\frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}$ and found that it has 0 valid solution(s) and 2 extraneous solution(s). In this section, we will address some common questions and concerns related to the equation and its solutions.

Q: What is the common denominator of the fractions in the equation?

A: The least common multiple (LCM) of $x-3$ and $x$ is $x(x-3)$.

Q: Why did we multiply both sides of the equation by the common denominator?

A: We multiplied both sides of the equation by the common denominator to eliminate the fractions and simplify the equation.

Q: How did we simplify the equation after multiplying both sides by the common denominator?

A: We expanded and simplified the equation by combining like terms.

Q: What is the quadratic equation that we obtained after simplifying the equation?

A: The quadratic equation is $x^2 - 4x + 2 = 0$.

Q: How did we solve the quadratic equation?

A: We used the quadratic formula to solve the equation.

Q: What are the two possible solutions to the quadratic equation?

A: The two possible solutions are $x = 2 + \sqrt{2}$ and $x = 2 - \sqrt{2}$.

Q: Why did we check the solutions to see if they satisfy the original equation?

A: We checked the solutions to ensure that they satisfy the original equation and are not extraneous solutions.

Q: What did we find when we checked the first solution?

A: We found that the first solution, $x = 2 + \sqrt{2}$, is an extraneous solution because the expression under the square root is undefined.

Q: What did we find when we checked the second solution?

A: We found that the second solution, $x = 2 - \sqrt{2}$, is also an extraneous solution because the expression under the square root is undefined.

Q: What is the final answer to the problem?

A: The equation has 0 valid solution(s) and 2 extraneous solution(s).

Q: Why is it essential to check the solutions to ensure that they satisfy the original equation?

A: It is essential to check the solutions to ensure that they satisfy the original equation because extraneous solutions can arise from the simplification process.

Q: What are some common mistakes to avoid when solving equations involving fractions?

A: Some common mistakes to avoid when solving equations involving fractions include:

• Not finding a common denominator
• Not multiplying both sides of the equation by the common denominator
• Not simplifying the equation correctly
• Not checking the solutions to ensure that they satisfy the original equation

Q: How can we ensure that we are solving equations correctly?

A: We can ensure that we are solving equations correctly by:

• Following the order of operations
• Simplifying the equation correctly
• Checking the solutions to ensure that they satisfy the original equation
• Avoiding common mistakes

Conclusion

In this Q&A article, we addressed some common questions and concerns related to the equation and its solutions. We emphasized the importance of checking the solutions to ensure that they satisfy the original equation and avoiding common mistakes when solving equations involving fractions. By following the steps outlined in this article, you can ensure that you are solving equations correctly and accurately.